+ Reply to Thread
Results 1 to 4 of 4

Thread: Different proportions in different conditions?

  1. #1
    Points: 319, Level: 6
    Level completed: 38%, Points required for next Level: 31

    Posts
    5
    Thanks
    1
    Thanked 0 Times in 0 Posts

    Different proportions in different conditions?




    Hi,

    I am struggling with the proper way to analyze my data. I did some research in and out of this forum but could not find the answer to my problem.

    I am looking at neurons and trying to analyse the proportion of protrusions (called spines) co-localizing with a protein S. So for each neuron, I have a number of S+ spines and a number of S- spines. I analyzed this in 14 control neurons and 15 neurons from my experimental condition.

    What I want to know is: Are the proportions of S- and S+ spines the same in my control and experimental conditions?

    You can see my data on the attached file below. Also, please note that the number of spines analyzed can be really different from one neuron to another (see last column of the attached file).

    I tried normalizing the data as percentages (percentages of S- spines and S+ spines for each neuron) but I am not sure I can do statistics after I normalized the data that way. Also it seems wrong that a neuron for which I analyzed 43 spines would have the same weight as the one I analyzed 600+ spines for.

    I also thought about using a binomial test on the raw data but I am not sure I can use my observed control data as "expected distribution". Can I? If so, that would mean I would have to pool the data from all neurons in each condition, would that be a problem?

    If anyone had some insight to share on how to analyze these data, I would greatly appreciate it. Thanks.
    Attached Images  

  2. #2
    Points: 319, Level: 6
    Level completed: 38%, Points required for next Level: 31

    Posts
    5
    Thanks
    1
    Thanked 0 Times in 0 Posts

    Re: Different proportions in different conditions?



    After some more thoughts and research, I understood that my variables are actually nominal variables and I should use contingency tables.
    I started with a contingency table that looked like that:

    ............S+ spine.....S- Spine
    Ctrl.........2413...........500
    Exp.........1556...........490

    I wanted to perform a Fisher test but Prism said my samples were too large and performed a Chi² test with Yates' correction instead. All seemed fine but when I went over Prism checklist for these two tests, I realized my variables might not really be independant (since several spines come from the same neurons). I tried looking up the Mantel-Haenszel test which seemed more appropriate. However I can't make one contingency table per neuron because those tables would only have one row.

    So even though I feel like I'm making progress towards the right way to analyze my data, I am still at a loss with my problem. It would really help if anyone was kind enough to give me some input on this.


    Edit: Still no one to help?...
    Last edited by Fanny-; 04-24-2017 at 02:39 AM. Reason: Desperate

  3. #3
    Points: 2,000, Level: 26
    Level completed: 40%, Points required for next Level: 150

    Location
    New Zealand
    Posts
    228
    Thanks
    3
    Thanked 48 Times in 47 Posts

    Re: Different proportions in different conditions?

    I think your question is whether there is a significant difference between the control and experiment for S- and for S+ (actually two questions).
    If the data was normal, two t tests would be fine - one for S- and one for S+. Unfortunately the data is not.
    I would try a permutation or other randomization test.

  4. #4
    TS Contributor
    Points: 12,501, Level: 73
    Level completed: 13%, Points required for next Level: 349
    rogojel's Avatar
    Location
    I work in Europe, live in Hungary
    Posts
    1,491
    Thanks
    162
    Thanked 334 Times in 314 Posts

    Re: Different proportions in different conditions?


    hi,
    I ran a permutation test in R and the p-value is around 0.15, so, no significant difference. This could BTW be due to the small sample size.

    Interestingly Moods Median also gave a comparable p value - around 0.2.


    Here is the code - maybe someone could double-check it?

    Code: 
    Len=500
    meandif=numeric(Len)
    df$simtype=rep("E",29)
    for(i in 1:Len){
      ind=sample(1:29, 14)
      df$simtype=rep("E",29)
      df$simtype[ind]="C"
      x=aggregate(df$percent, list(df$simtype), mean)
      meandif[i]=x$x[2]-x$x[1]
    }
    hist(meandif)
    p=sum(meandif>0.0378)/Len
    p
    The magical 0.0378 is the actual measured difference between the control and experimental group.

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats